In the given figure, sides QP and RQ of ΔPQR are produced to points S and T respectively. If ∠SPR = 135º and ∠PQT = 110º, find ∠PRQ.
In the given figure, sides QP and RQ of ΔPQR are

Solution:

It is given that,

$\angle S P R=135^{\circ}$ and $\angle P Q T=110^{\circ}$

$\angle S P R+\angle Q P R=180^{\circ}($ Linear pair angles $)$

$\Rightarrow 135^{\circ}+\angle Q P R=180^{\circ}$

$\Rightarrow \angle Q P R=45^{\circ}$

Also, $\angle \mathrm{PQT}+\angle \mathrm{PQR}=180^{\circ}$ (Linear pair angles)

$\Rightarrow 110^{\circ}+\angle \mathrm{PQR}=180^{\circ}$

$\Rightarrow \angle \mathrm{PQR}=70^{\circ}$

As the sum of all interior angles of a triangle is $180^{\circ}$, therefore, for $\triangle \mathrm{PQR}$,

$\angle \mathrm{QPR}+\angle \mathrm{PQR}+\angle \mathrm{PRQ}=180^{\circ}$

$\Rightarrow 45^{\circ}+70^{\circ}+\angle \mathrm{PRQ}=180^{\circ}$

$\Rightarrow \angle \mathrm{PRQ}=180^{\circ}-115^{\circ}$

$\Rightarrow \angle \mathrm{PRQ}=65^{\circ}$
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